Intergration by substitution

**camherokid** · July 1st 2007, 11:35 AM

Do do you this one please

**red_dog** · July 1st 2007, 11:41 AM

$\displaystyle \int\frac{1+x}{1+x^2}dx=\int\frac{1}{1+x^2}dx+\int \frac{x}{1+x^2}dx=$
$\displaystyle =\arctan x+\frac{1}{2}\int\frac{2x}{1+x^2}dx=\arctan x+\frac{1}{2}\ln (1+x^2)+C$

**Jhevon** · July 1st 2007, 11:43 AM

Originally Posted by camherokid

Do do you this one please

$\int \frac {1 + x}{1 + x^2}~dx = \int \frac {1}{1 + x^2}~dx + \int \frac {x}{1 + x^2}~dx$

$= \arctan x + \int \frac {x}{1 + x^2}~dx$

We proceed by substitution for the second integral:

Let $u = 1 + x^2$

$\Rightarrow du = 2x ~dx$

$\Rightarrow \frac {1}{2} du = x ~dx$

$\Rightarrow \int \frac {1 + x}{1 + x^2}~dx = \arctan x + \frac {1}{2} \int \frac {1}{u} ~du$

$= \arctan x + \frac {1}{2} \ln u + C$

$= \boxed { \arctan x + \frac {1}{2} \ln \left( 1 + x^2 \right) + C }$

EDIT: You're too fast for me red_dog!

**red_dog** · July 1st 2007, 11:44 AM

$f(x)=\tan ^3x$ is an odd function and $\displaystyle \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$ is a symetric interval, so the integral is $0$ .

**Jhevon** · July 1st 2007, 11:52 AM

I will do the indefinite integral for the second, i leave plugging in the limits to you

$\int \tan^3 x ~dx = \int \tan^2 x \tan x ~dx$

$= \int \left( \sec^2 x - 1 \right) \tan x ~dx$

$= \int \left( \sec^2 x \tan x - \tan x \right)~dx$

$= \int \sec^2 x \tan x ~dx - \int \tan x ~dx$

$= \int \sec^2 x \tan x ~dx - \ln | \sec x | + C$

We proceed by substitution for the first integral:

Let $u = \tan x$

$\Rightarrow du = \sec^2 x ~dx$

$\Rightarrow \int \tan^3 x ~dx = \int u ~du - \ln | \sec x | + C$

$= \frac {1}{2} u^2 - \ln | \sec x | + C$

$= \boxed { \frac {1}{2} \tan^2 x - \ln | \sec x | + C }$

**Jhevon** · July 1st 2007, 11:54 AM

Originally Posted by red_dog

$f(x)=\tan ^3x$ is an odd function and $\displaystyle \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$ is a symetric interval, so the integral is $0$ .

Ah, i never noticed that (never paid attention to the limits, i planned on finding the indefinate integral from the start). That's true, the integral is 0. However, if what you actually wanted was the area, you would find $2 \int_{0}^{ \frac { \pi}{6}} \tan^3 x~dx$

**galactus** · July 1st 2007, 02:13 PM

Another possible solution, similar to Jhevon's is:

$\int{tan^{3}(x)}dx=\int{(sec^{2}(x)-1)tan(x)}dx=sec^{2}(x)tan(x)-tan(x)$

$\int{sec^{2}(x)tan(x)}dx$

Now, let $u=sec(x), \;\ du=sec(x)tan(x)dx$

$\int{u}du=\frac{1}{2}u^{2}=\frac{1}{2}sec^{2}(x)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\int{tan(x)}dx$

$\int\frac{sin(x)}{cos(x)}dx$

Let $u=cos(x), \;\ -du=sin(x)dx$

$-\int\frac{1}{u}du=ln(u)=-ln(cos(x))$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So, we have $\boxed{\frac{1}{2}sec^{2}(x)+ln(cos(x))+C}$

$\frac{1}{2}tan^{2}(x)-ln(sec(x))+\frac{1}{2}=\frac{1}{2}sec^{2}(x)+ln(co s(x))$

**curvature** · July 2nd 2007, 04:28 AM

Originally Posted by red_dog

$f(x)=\tan ^3x$ is an odd function and $\displaystyle \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$ is a symetric interval, so the integral is $0$ .

This is the simplest method of all.
The integral of an odd function on a symmetric interval is always zero

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